Difference between revisions of "2013 AMC 8 Problems/Problem 18"

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==Solution==
 
==Solution==
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There are <math>10 \cdot 12 = 120</math> cubes on the base of the box. Then, for each of the 4 layers above the bottom (as since each cube is 1 foot by 1 foot by 1 foot and the box is 5 feet tall, there are 4 feet left), there are <math>9 + 11 + 9 + 11 = 40</math> cubes. Hence, the answer is <math>120 + 4 \cdot 40 = \boxed{\textbf{(B)}\ 280}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2013|num-b=17|num-a=19}}
 
{{AMC8 box|year=2013|num-b=17|num-a=19}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 16:56, 27 November 2013

Problem

Isabella uses one-foot cubical blocks to build a rectangular fort that is 12 feet long, 10 feet wide, and 5 feet high. The floor and the four walls are all one foot thick. How many blocks does the fort contain?

[asy here]

$\textbf{(A)}\ 204 \qquad \textbf{(B)}\ 280 \qquad \textbf{(C)}\ 320 \qquad \textbf{(D)}\ 340 \qquad \textbf{(E)}\ 600$

Solution

There are $10 \cdot 12 = 120$ cubes on the base of the box. Then, for each of the 4 layers above the bottom (as since each cube is 1 foot by 1 foot by 1 foot and the box is 5 feet tall, there are 4 feet left), there are $9 + 11 + 9 + 11 = 40$ cubes. Hence, the answer is $120 + 4 \cdot 40 = \boxed{\textbf{(B)}\ 280}$.

See Also

2013 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 17
Followed by
Problem 19
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AJHSME/AMC 8 Problems and Solutions

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